Solve each equation. $$ 81^{-2 b-1} \cdot 81^{-2 b-1}=\left(\frac{1}{9}\right)^{2 b} $$

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Solve the equation $$ 81^{-2b-1} \cdot 81^{-2b-1} = \left(\frac{1}{9}ight)^{2b} $$.
Solve the equation by following steps:
- step0: Solve for $$b$$:
  $$81^{-2b-1}	imes 81^{-2b-1}=\left(\frac{1}{9}ight)^{2b}$$
- step1: Multiply the terms:
  $$81^{-4b-2}=\left(\frac{1}{9}ight)^{2b}$$
- step2: Rewrite the expression:
  $$9^{-8b-4}=9^{-2b}$$
- step3: Set the exponents equal:
  $$-8b-4=-2b$$
- step4: Move the variable to the left side:
  $$-8b-4+2b=0$$
- step5: Add the terms:
  $$-6b-4=0$$
- step6: Move the constant to the right side:
  $$-6b=0+4$$
- step7: Remove 0:
  $$-6b=4$$
- step8: Change the signs:
  $$6b=-4$$
- step9: Divide both sides:
  $$\frac{6b}{6}=\frac{-4}{6}$$
- step10: Divide the numbers:
  $$b=-\frac{2}{3}$$
The solution to the equation $$81^{-2b-1} \cdot 81^{-2b-1} = \left(\frac{1}{9}ight)^{2b}$$ is $$b = -\frac{2}{3}$$ or $$b = -0.\dot{6}$$.
$$b = -\frac{2}{3}$$ or $$b = -0.\dot{6}$$

Solve each equation.

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